Is $N O^{-}$ $NO^-$ Paramagnetic or Diamagnetic?

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Is $N O^{-}$ $NO^-$ Paramagnetic or Diamagnetic?

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Answer 1 · 2016-04-07T19:26:51

The MO diagram for $NO$ $"NO"$ is as follows (Miessler et al., Answer Key):

(The original was this; I added the orbital depictions and symmetry labels.)

Quick overview of what the labels correspond to what MOs:

$1 a_{1}$ $1a_1$ is the $σ_{2 s}$ $sigma_(2s)$ bonding MO.
$2 a_{1}$ $2a_1$ is the $σ_{2 s}^{*}$ $sigma_(2s)^"*"$ antibonding MO.
$1 b_{1}$ $1b_1$ is the $π_{2 p_{x}}$ $pi_(2p_x)$ bonding MO.
$1 b_{2}$ $1b_2$ is the $π_{2 p_{y}}$ $pi_(2p_y)$ bonding MO.
$3 a_{1}$ $3a_1$ is the $σ_{2 p_{z}}$ $sigma_(2p_z)$ bonding MO, but it's relatively nonbonding with respect to oxygen.
$2 b_{1}$ $2b_1$ is the $π_{2 p_{x}}^{*}$ $pi_(2p_x)^"*"$ antibonding MO.
$2 b_{2}$ $2b_2$ is the $π_{2 p_{y}}^{*}$ $pi_(2p_y)^"*"$ antibonding MO.
$4 a_{1}$ $4a_1$ is the $σ_{2 p_{z}}^{*}$ $sigma_(2p_z)^"*"$ antibonding MO.

Since this is $NO$ $"NO"$ , if you add an electron to get to ${NO}^{-}$ $"NO"^(-)$ , you add it into the $2 b_{2}$ $2b_2$ orbital, which is the $\mathbf(pi_(2p_y)^"*")$ antibonding MO.

That increases its paramagnetic properties, as there exist two unpaired electrons now instead of just one.

CHALLENGE: What does that do to the $N-O$ $pi$ bond? Does it weaken or strengthen it? (Hint: Consider the bond order). What about $NO^(+)$ ? Is that diamagnetic, and how do you know?

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Is $N O^{-}$ $NO^-$ Paramagnetic or Diamagnetic?

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